JEE Advanced | 2013ThermochemistryHard

Question

The standard enthalpies of formation of CO2(g), H2O(l) and glucose(s) at 250C are -400 kJ/mol, -300 kJ/mol & -1300 kJ/mol respectively. The standard enthalpy of combustion per gram of glucose at 250C is -

Options

A.+ 2900 kJ
B.- 2900 kJ
C.- 16.11 kJ
D.+ 16.11 kJ

Solution

Combustion reaction for glucose is
C6H12O6(s) + 6O2(g)→ 6CO2(g) + 6H2O(l)
ᐃHr = 6(-400) + 6(-300) - (-1300)
= -2900 KJ/mol KJ / gm
= -16.11 KJ/gm

Create a free account to view solution

View Solution Free
Topic: Thermochemistry·Practice all Thermochemistry questions

More Thermochemistry Questions

Enthalpy of neutralisation of CH3COOH by NaOH is - 50.6 kJ/mol and the heat of neutralisation of a strong acid with NaOH...When 1 g equivalent of strong acid reacts with strong base, the heat released is 13.5 kcal. When 1 g equivalent H2A is c...The enthalpy change when x g of phenol dissolves in y g of CHCl3 is given below at a certain temperature.x Y ΔH (kcal)0....The lattice energy of solid NaCl is 180 kcal/mol. The dissolution of the solid in water, in the form of ions is endother...2MnO4 – + 16H+ + 10Cl– → 2Mn2+ + 5Cl2(g) + 8H2OThe above reaction is endothermic and hence, the actual temperature of th...